# How to show the following equalities as suggested by Wolfram Alpha? [closed]

I am not very experienced at solving inequalities and I cannot understand the following: I was given the inequality

$$\tag{1} 2xy>x^2-y^2.$$

When I put this into WolframAlpha I get $$y>\sqrt2 \sqrt{x^2} - x \text{ and } y<-\sqrt2 \sqrt{x^2} - x$$

How do I come to these two inequalities from $(1)$?

## closed as off-topic by user21820, Dmoreno, Mostafa Ayaz, callculus, Ethan BolkerFeb 21 '18 at 17:53

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[Edited after typos in OP are corrected.]

When I ask WolframAlpha I get the answer $y>\sqrt2 \sqrt{x^2} - x$ and $y<-\sqrt2 \sqrt{x^2} - x$. How do I find this solution?

WolframAlpha shows you:

To see how the results are found, taking $x$ as a constant and solving the quadratic equation in $y$: $$y^2+2xy-x^2=0\tag{*}$$ one has $$\frac{-2x\pm\sqrt{4x^2+4x^2}}{2}=-x\pm\sqrt{2}\sqrt{x^2}=-x\pm\sqrt{2}|x|.$$ One can recognize these two roots in the "Results" above.

Note that the two roots can be written as (regardless $x$ is positive or not) $$y_1=(-1+\sqrt{2})x,\quad y_2=(-1-\sqrt{2})x.$$

• When $x\geqslant 0$, $y_1\geqslant y_2$ and the inequality $y^2+2xy-y^2>0$, which is equivalent to $(y-y_1)(y-y_2)>0$, implies that $$y>y_1\quad \hbox{or}\quad y<y_2.$$

• When $x<0$, $y_1<y_2$. Now the inequality $y^2+2xy-y^2>0$ implies that $$y>y_2\quad \hbox{or}\quad y<y_1.$$

Combining these two cases together, one can see that for $x\in{\bf R}$, $$y<-x-\sqrt{2}|x|\qquad\hbox{or}\quad y>-x+\sqrt{2}|x|.$$

• Alright I think I understand this answer. Thank you! I also know that x>0 and y>0, what impact will this have on the answer? The last case is not true I suppose? What about when x>y? I agree about the negative votes, very strange! I wonder why my question was downvoted so much. – Emilia314 Sep 30 '17 at 18:16
• @Mathmellow Solving the inequality $2xy>x^2-y^2$ really means finding the set $S=\{(x,y)\in{\bf R}^2\mid 2xy>x^2-y^2\}$. Note that WolframAlpha tells you that $S=\{(x,y)\in{\bf R}^2\mid y<-x-\sqrt{2}|x|\ \hbox{ or}\ y>-x+\sqrt{2}|x|\}$ (and you can see the picture in Will Jay's answer.) If you further assume that $x>0$ and $y>0$, the solution set would become $S\cap Q$ where $Q=\{(x,y)\mid x>0\ \hbox{and }y>0\}$, similarly for other constraints. Does this answer your question? – Jack Sep 30 '17 at 18:22
• Yes, I think so. Will's answer was also very intuitive and helpful. May I type up my answer here as well so other people may comment on it, in case I misunderstood? – Emilia314 Sep 30 '17 at 18:47
• @Mathmellow: you could certainly post your own answer. – Jack Sep 30 '17 at 18:59

Hint:

If $y=0$, we have $x^2<0$, which is impossible for real $x$.

So dividing both sides by $y^2$, we get $$\left(\dfrac xy\right)^2-2\left(\dfrac xy\right)-1<0.$$

On the other hand, for $(z-a)(z-b)<0$ with $a<b$, we can prove that $$a<z<b.$$

• What are $a$ and $b$? – Jack Sep 30 '17 at 16:22
• @Jack The roots of the quadratic Lab decscribed – Jyrki Lahtonen Sep 30 '17 at 16:26
• @JyrkiLahtonen: which quadratic equation are you referring to? – Jack Sep 30 '17 at 16:27
• @Jack $$\left(\frac xy\right)^2-2\left(\frac xy\right)-1.$$ – Jyrki Lahtonen Sep 30 '17 at 16:29
• @Jack then shouldn't it be $(\frac{x}{y}-a)(\frac{x}{y}-b)<0$? Exactly! That's Lab's point, I think. He was just describing the set of solutions of a quadratic inequality. That he used $x/y$ as the variable in one and $x$ in the other is inconsequential. The ratio $x/y$ is what shows in this problem, the other was a generic quadratic. – Jyrki Lahtonen Sep 30 '17 at 16:48

You want $y^2 + 2 yx - x^2 > 0.$ The boundaries are $y^2 + 2 yx - x^2 = 0.$ Let the slope of a line be $$m = \frac {y}{x},$$ divide $y^2 + 2 yx - x^2 = 0$ by $x^2$ for nonzero $x,$ giving $$m^2 + 2m - 1 = 0.$$

Put another way, you want $$\left( y - \left( \sqrt 2 - 1 \right) x \right) \left( y + \left( \sqrt 2 + 1 \right) x \right) > 0$$ Either both factors are positive or both negative. We can see which of the four quarters in the picture work by noticing that $(0,2)$ works, and $(0,-2)$ works. The borders are the depicted slanted lines. The slanted lines are called the null cone of the (indefinite) quadratic form $y^2 + 2 yx - x^2.$

$2xy\gt x^2-y^2;$

$2y/x \gt 1 - (y/x)^2$.

$[y/x +1]^2 \gt 2.$

$(y+x)^2 \gt 2x^2.$

1)$y+x \ge 0: y\ge -x.$

$y+x \gt \sqrt{2x^2}.$

$y \gt \sqrt{2x^2} -x.$

2) $y+x \lt 0; y \lt -x.$

$-(y+x) \gt \sqrt{2x^2}$.

$y \lt -\sqrt{2x^2} - x$.